Problem: If $f(3)=1$ and $f(2x)=2f(x)$ for all $x$, find $f^{-1}(64)$.
We are looking for some $x$ such that $f(x)=64$.  We notice that by doubling $x$ we can double $f(x)$ as well and also that $f(3)=1$.

Applying $f(2x)=2f(x)$ repeatedly, we have: \begin{align*}
f(3)&=1,\\
f(6)&=2,\\
f(12)&=4,\\
f(24)&=8,\\
f(48)&=16,\\
f(96)&=32,\\
f(192)&=64.
\end{align*}So $f^{-1}(64)=\boxed{192}$.